The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was . On adding 0.1 mo — Solutions and Colligative Properties Chemistry Question
Question
The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was $-0.80^\circ\text{C}$. On adding 0.1 mole of $\text{Hg(CN)}_2$ in the solution containing 1 kg of water, the freezing point of the solution was $-0.6^\circ\text{C}$. Assuming that the complex is formed according to the equation: $\text{Hg(CN)}_2 + m \text{ CN}^- \rightarrow \text{Hg(CN)}_{m+2}^{m-}$ and $\text{Hg(CN)}_2$ is the limiting reactant, the value of $m$ is
💡 Solution & Explanation
Initial effective molality: $\Delta T_{f1} = 0.80 = 1.86 \times m_{eff1} \implies m_{eff1} = 0.80 / 1.86$. Final effective molality: $\Delta T_{f2} = 0.60 = 1.86 \times m_{eff2} \implies m_{eff2} = 0.60 / 1.86$. Thus, $m_{eff2} / m_{eff1} = 0.60 / 0.80 = 3/4$. Initially, KCN gives 0.2 $\text{K}^+$ and 0.2 $\text{CN}^-$, so $m_{eff1} = 0.40$ (assuming theoretical complete dissociation, actual $i$ is slightly less, but ratio holds). Adding 0.1 mole $\text{Hg(CN)}_2$ consumes $0.1m$ moles of $\text{CN}^-$ and forms 0.1 moles of complex. Final particles = $0.2 (\text{K}^+) + (0.2 - 0.1m) (\text{CN}^-) + 0.1 (\text{complex}) = 0.5 - 0.1m$. Setting $(0.5 - 0.1m) / 0.40 = 3/4 \implies 0.5 - 0.1m = 0.3 \implies 0.1m = 0.2 \implies m = 2$. Therefore, correct answer is 2.